Shortcut in Standard Deviation questions???

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This question is already solved on this forum and I know the solution also. But can anyone tell me the shortcut to solve this kind of problem. May be some properties of SD I am missing...

Thanks in advance.

I have calculated SD for each answers choices over here and (D) is closest to the SD of given 5 numbers.

Question:

A set of data consists of the following 5 numbers: 0,2,4,6, and 8. Which two numbers, if added to create a set of 7 numbers, will result in a new standard deviation that is close to the standard deviation for the original 5 numbers?

A). -1 and 9
B). 4 and 4
C). 3 and 5
D). 2 and 6
E). 0 and 8

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by rs007 » Mon May 05, 2008 10:04 am
working with the 5 numbers we find that the mean = 4.
First compute the std dev of the 5 numbers (40/5=8)

also scanning through the answer choices gives us mean = 4 as well since each number is equidistant (in either direction of 4)

Once we know that, we can proceed with the following:

let x be the sum of squares of difference of the two unknown numbers.
(40 +x)/7 = 8 (if we want it to be exactly the same as the std dev of the 5 numbers)

solving x = 16. So we have the find the sum of square of differences of the answer choices from the mean 4 that is closest to 16

this gives you D.

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by VP_RedSoxFan » Mon May 05, 2008 3:05 pm
Codesnooker,

If you're looking for a shortcut to this problem, let's reason through the properties of SD together get the answer without any computation.

The question asks which two numbers when added to the set give a new stdev that is closest to the original stdev of 5 numbers.

rs007 makes a great observation that all of the answer choices have mean 4 like the original set of 5 so what we're really looking to do is look at the effect of adding the answer choices to the set on the dispersion of the items in the set.

[A]: -1 and 9: These two numbers are farther away from 4 than any in the original set so clearly it will make the stdev bigger; 5 observations closer to mean, 0 farther

: 4 and 4: These two numbers are as close to the mean as possible so adding them will shrink the stdev; 4 observations farther from mean, 1 closer (or as close)

[C]: 3 and 5: Adding these two are good candidates because they are "in the middle" of the set. However, they are too close to the mean compared to another answer choice; 4 observations farther from mean, 1 closer

[D]: 2 and 6: Adding these two is the correct answer because they would yield where they didn't unduly influence the overall dispersion of the set; 2 farther from mean, 1 closer, 2 as close

[E]: 0 and 8: see comments for [A]

Basically, without calculating, you're looking for the two numbers that, when added to the set, affect the stdev (overall dispersion from mean) of the set the least.
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by codesnooker » Mon May 05, 2008 11:01 pm
Thanks rs007, I got your point.
Thanks Ron, through I you I come to know how to eliminate the most of the choice.

However, I am not completely agree when I look at choices C and D, as they are very close and we may hit the wrong option at the end. So I would prefer rs007 solution only for these 2 choices after eliminating the other one by your technique.

However, in lack of time your technique is quite efficient. May be I need more practice on such question to grasp the concept explained by you.

Thanks to both of you.

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by resilient » Wed Jun 25, 2008 4:06 am
the puzzling part here is that when counting the satandard deviations choice c d and e all count up to 8 MOVEMENTS from the mean. I think that we are getting hung up on this part.
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by saege » Sat Jun 28, 2008 8:28 am
VP_RedSoxFan wrote:Codesnooker,

If you're looking for a shortcut to this problem, let's reason through the properties of SD together get the answer without any computation.

The question asks which two numbers when added to the set give a new
[A]: -1 and 9: These two numbers are farther away from 4 than any in the original set so clearly it will make the stdev bigger; 5 observations closer to mean, 0 farther

: 4 and 4: These two numbers are as close to the mean as possible so adding them will shrink the stdev; 4 observations farther from mean, 1 closer (or as close)

[C]: 3 and 5: Adding these two are good candidates because they are "in the middle" of the set. However, they are too close to the mean compared to another answer choice; 4 observations farther from mean, 1 closer

[D]: 2 and 6: Adding these two is the correct answer because they would yield where they didn't unduly influence the overall dispersion of the set; 2 farther from mean, 1 closer, 2 as close



Red Sox Can you explain what you meant by your statements. I have highlighted them in red . Thanks in advance.

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by axat » Mon Jul 20, 2009 5:08 am
VP_RedSoxFan, I am unable to understand your explaination.

I would request you and other instructors(Stacey, Ian etc) to elaborate on the different approaches to solve this question. I know that this is a question that can be solved without any lengthy calculations and purely on the basis of the principles of dispersion.

I await your respective responses.


VP_RedSoxFan wrote:Codesnooker,

If you're looking for a shortcut to this problem, let's reason through the properties of SD together get the answer without any computation.

The question asks which two numbers when added to the set give a new stdev that is closest to the original stdev of 5 numbers.

rs007 makes a great observation that all of the answer choices have mean 4 like the original set of 5 so what we're really looking to do is look at the effect of adding the answer choices to the set on the dispersion of the items in the set.

[A]: -1 and 9: These two numbers are farther away from 4 than any in the original set so clearly it will make the stdev bigger; 5 observations closer to mean, 0 farther

: 4 and 4: These two numbers are as close to the mean as possible so adding them will shrink the stdev; 4 observations farther from mean, 1 closer (or as close)

[C]: 3 and 5: Adding these two are good candidates because they are "in the middle" of the set. However, they are too close to the mean compared to another answer choice; 4 observations farther from mean, 1 closer

[D]: 2 and 6: Adding these two is the correct answer because they would yield where they didn't unduly influence the overall dispersion of the set; 2 farther from mean, 1 closer, 2 as close

[E]: 0 and 8: see comments for [A]

Basically, without calculating, you're looking for the two numbers that, when added to the set, affect the stdev (overall dispersion from mean) of the set the least.

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by Ian Stewart » Mon Jul 20, 2009 10:06 am
axat wrote: I would request you and other instructors(Stacey, Ian etc) to elaborate on the different approaches to solve this question. I know that this is a question that can be solved without any lengthy calculations and purely on the basis of the principles of dispersion.
It can't, really. It's not a realistic GMAT question - it's certainly not an official question. If you have a good intuition about standard deviation -- if, say, you're a professional statistician -- you can likely 'eyeball' the answer here, but that's far more than is expected of the GMAT test taker. The difficulty here is that the size of the set changes when we add new elements, so you'd need to decide how the larger denominator in the averaging part of the standard deviation calculation balances with the additional distances to the mean from the newly added elements. To be sure of the answer, you'd need to use the standard deviation formula, which no real GMAT question requires. Even the correct answer to the question above changes the standard deviation a fair bit, so it's not especially obvious (to leave the standard deviation unchanged in the question above, you'd want to add the elements 4 - 2*root(2) and 4 + 2*root(2), which are roughly 1.2 and 6.8, so you can see that (0,8) is close to being the right answer here).

I've found that many prep company questions about standard deviation do not resemble real GMAT questions very closely. To get a good idea of what you're actually expected to know about standard deviation on the real test, it's best to consult official questions.
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